Let m<n. We want to show that $\displaystyle \|f_n-f_m\|_2\to0$ as $\displaystyle m,\,n\to\infty$, where $\displaystyle \|f_n-f_m\|_2^2 = \int_0^2(f_n(x)-f_m(x))^2\,dx$. But $\displaystyle f_m(x)$ and $\displaystyle f_n(x)$ coincide on the intervals [0,1-(1/m)] and [1,2]. So we need only worry about the integral on the interval from 1-(1/m) to 1. Even on this interval, the integral is a bit messy to evaluate exactly, so I suggest using a little trick.

It's easy to see that if m<n then $\displaystyle f_m(x)\geqslant f_n(x)$. Therefore $\displaystyle 0\leqslant f_m(x) - f_n(x) \leqslant f_m(x)\leqslant1$, and the integral that we want to evaluate is less than $\displaystyle \int_{1-(1/m)}^11\,dx$. That integral is (very!) easy to evaluate, and it gets small as m gets large.